Question

Let \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} , \ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1. \) If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (I + A)^8 \) is:

• A. \( \begin{bmatrix} 4 & -1 \\6 & -1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 257 & -64 \\514 & -127 \end{bmatrix} \)
• C. \( \begin{bmatrix} 1025 & -511 \\2024 & -1024 \end{bmatrix} \)
• D. \( \begin{bmatrix} 766 & -255 \\1530 & -509 \end{bmatrix} \)

Hint:

When a matrix satisfies \( A^2 = A \), powers simplify: \( A^n = A \). Use this in binomial expansions.

Answer 1

The Correct Option is D

This problem necessitates the initial determination of the values for \( \alpha \) and \( \beta \) based on the provided conditions for matrix \( A \). Following the determination of matrix \( A \), the subsequent task is to compute the matrix \( (I + A)^8 \).

Concept Used:

1. Determinant of a 2x2 Matrix: The determinant of a matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated as \( \det(M) = ad - bc \).

2. System of Equations: Solving for two unknown variables requires two independent equations.

3. Matrix Exponentiation via Diagonalization: If a matrix \( B \) can be expressed as \( B = PDP^{-1} \), where \( D \) is a diagonal matrix containing eigenvalues and \( P \) is the matrix of corresponding eigenvectors, then \( B^n = PD^n P^{-1} \). Eigenvalues \( \lambda \) are derived from the characteristic equation \( \det(B - \lambda I) = 0 \).

Step-by-Step Solution:

The matrix \( A \) is given as \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} \), with the constraints \( \alpha > 0 \), \( \det(A) = 0 \), and \( \alpha + \beta = 1 \).

The determinant of \( A \) is computed as follows:

\[ \det(A) = (\alpha)(\beta) - (-1)(6) = \alpha\beta + 6 \]

Applying the condition \( \det(A) = 0 \), we obtain:

\[ \alpha\beta + 6 = 0 \implies \alpha\beta = -6 \]

We now have a system of two linear equations with two variables:

\[ \alpha + \beta = 1 \] \[ \alpha\beta = -6 \]

This system can be solved by considering a quadratic equation \( t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0 \), where \( \alpha \) and \( \beta \) are the roots. This leads to:

\[ t^2 - (1)t + (-6) = 0 \implies t^2 - t - 6 = 0 \]

Factoring the quadratic equation yields:

\[ (t - 3)(t + 2) = 0 \]

The roots are \( 3 \) and \( -2 \). Given that \( \alpha > 0 \), we assign \( \alpha = 3 \) and \( \beta = -2 \).

Consequently, the matrix \( A \) is:

\[ A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \]

The objective is to compute \( (I + A)^8 \). First, we determine the matrix \( B = I + A \):

\[ B = I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} \]

To calculate \( B^8 \), we employ diagonalization. The eigenvalues of \( B \) are found by solving the characteristic equation \( \det(B - \lambda I) = 0 \):

\[ \det\left(\begin{bmatrix} 4-\lambda & -1 \\ 6 & -1-\lambda \end{bmatrix}\right) = (4-\lambda)(-1-\lambda) - (-1)(6) = 0 \] \[ -4 - 4\lambda + \lambda + \lambda^2 + 6 = 0 \] \[ \lambda^2 - 3\lambda + 2 = 0 \] \[ (\lambda - 1)(\lambda - 2) = 0 \]

The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).

The corresponding eigenvectors are determined as follows. For \( \lambda_1 = 1 \):

\[ (B - 1I)\mathbf{v}_1 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 3x - y = 0 \]

An eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \).

For \( \lambda_2 = 2 \):

\[ (B - 2I)\mathbf{v}_2 = \begin{bmatrix} 2 & -1 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 2x - y = 0 \]

An eigenvector is \( \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \).

The matrices \( P \), \( D \), and \( P^{-1} \) are constructed as:

\[ P = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \] \[ P^{-1} = \frac{1}{(1)(2) - (1)(3)} \begin{bmatrix} 2 & -1 \\ -3 & 1 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 2 & -1 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} \]

Final Computation & Result:

The formula \( B^8 = PD^8P^{-1} \) is applied. First, \( D^8 \) is computed:

\[ D^8 = \begin{bmatrix} 1^8 & 0 \\ 0 & 2^8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} \]

Next, \( B^8 \) is calculated:

\[ B^8 = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} \]

The product of the first two matrices is:

\[ \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} = \begin{bmatrix} 1 & 256 \\ 3 & 512 \end{bmatrix} \]

Multiplying this result by \( P^{-1} \) yields:

\[ B^8 = \begin{bmatrix} 1 & 256 \\ 3 & 512 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} (1)(-2) + (256)(3) & (1)(1) + (256)(-1) \\ (3)(-2) + (512)(3) & (3)(1) + (512)(-1) \end{bmatrix} \] \[ B^8 = \begin{bmatrix} -2 + 768 & 1 - 256 \\ -6 + 1536 & 3 - 512 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix} \]

Therefore, the matrix \( (I+A)^8 \) is \( \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix} \).